find the measure of each segment

Find The Measure Of Each Segment

Answers

Answer 1

The value of each line segment is found as 26 units.

What is defined as the mid point?A midpoint is a point in the center of a line connecting two points. The two points of reference are the line's endpoints, and the midpoint is located between the two. The midpoint splits the line connecting such two points in half. Furthermore, a line drawn to bisect the line connecting these two points passes through midpoint.

For the given question.

The line is given as DE and the mid point of line is D.

Thus,

CD = DE   .....eq 1

The value of each term are given as;

CD = 2x + 7

DE = 4(x - 3)

Put the value in equation 1.

2x + 7 = 4(x - 3)

2x + 7 = 4x - 12

Bring variables and constants on the different sides.

4x - 2x = 7 + 12

2x = 19

x = 9.5

Put the value in each side;

CD = 2×9.5 + 7 = 26 unitsDE = 4(9.5 - 3) = 26 units

Thus, the value of each line segment is found as 26 units.

To know more about the mid point, here

https://brainly.com/question/12223533

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Related Questions

A box of a granola contains 16.8 ounces . It cost $5.19 . What is the cost , to the nearest cent , of the granola per ounce ? A . $0.12 B . $0.31 C . $3.24

Answers

The cost per unit ounce is obtained by computing the quotient:

[tex]c=\frac{C}{N}.[/tex]

Where:

• c is the cost per unit ounce,

,

• C is the cost,

,

• N is the number of ounces that you get for C.

In this problem we have:

• C = $5.19,

,

• N = 16.8 ounces.

Computing the quotient, we get:

[tex]c=\frac{5.19}{16.8}\cong0.31[/tex]

dollars per ounce.

Answer: B. $0.31

A building is 5 feet tall. the base of the ladder is 8 feet from the building. how tall must a ladder be to reach the top of the building? explain your reasoning.show your work. round to the nearest tenth if necessary.

Answers

The ladder must be 9.4 ft to reach the top of the building

Here, we want to get the length of the ladder that will reach the top of the building

Firstly, we need a diagrammatic representation

We have this as;

As we can see, we have a right triangle with the hypotenuse being the length of the ladder

We simply will make use of Pythagoras' theorem which states that the square of the hypotenuse is equal to the sum of the squares of the two other sides

Thus, we have;

[tex]\begin{gathered} x^2=5^2+8^2 \\ x^2=\text{ 25 + 64} \\ x^2\text{ = 89} \\ x=\text{ }\sqrt[]{89} \\ x\text{ = 9.4 ft} \end{gathered}[/tex]

solve the equation by completing the square. Show all solutions8x^2 + 16x = 42

Answers

8x² + 16x = 42​

x² + 16/8x = 42​/8 dividing by 8 at both sides

x² + 2x = 5.25

x² + 2x - 5.25 = 0

If we compute (x + 1)², we get:

(x + 1)² = x² + 2*x*1 + 1² = x² + 2x + 1

Then,

x² + 2x - 5.25 + 1 - 1 = 0

(x² + 2x + 1) + (-5.25 - 1) = 0

(x + 1)² - 6.25 = 0

(x + 1)² = 6.25

x + 1 = √6.25

This has 2 solutions,

x + 1 = 2.5 or x + 1 = -2.5

x = 2.5 - 1 x = -2.5 - 1

x = 1.5 x = -3.5

Determine (Freshman) Small Cafeteria). Interpret this answer in the context of the situation.

Answers

Step 1:

[tex]\text{Probability = }\frac{N\text{umber of required outcomes}}{N\text{umber of total possible outcome}}[/tex]

Step 2:

a)

Total possible outcome = 2640

Total number of freshman = 625

[tex]\begin{gathered} P(\text{Freshman) = }\frac{625}{2640} \\ \text{= }\frac{125}{528} \\ \text{= 0.237} \end{gathered}[/tex]

Step 3:

b)

Total number of senior and large cafeteria = 350

[tex]\begin{gathered} P(\text{senior and large cafeteria) = }\frac{350}{2640} \\ =\text{ }\frac{70}{528} \\ =\text{ }\frac{35}{264} \\ =\text{ 0.132} \end{gathered}[/tex]

Step 4:

c)

Number of Sophomore or student center = 650 + 595 - 125 = 1120

[tex]\begin{gathered} P(\text{Sophomore or student center) = }\frac{1120}{2640} \\ =\text{ }\frac{112}{264} \\ =\text{ 0}.424 \end{gathered}[/tex]

Step 5:

d)

[tex]\begin{gathered} p(\text{freshman}|\text{small cafeteria) = }\frac{n(freahman\text{ and small cafeteria)}}{n(small\text{cafeteria)}} \\ =\frac{435}{860}\text{ } \\ =\text{ }\frac{87}{172} \\ =\text{ 0.506} \end{gathered}[/tex]

review the rental and purchase property information to answer the question: calculate the difference in total move-in cost between the two properties. $31,497.35 $35,842.95$39,285.45$4,976.55

Answers

Let us calculate the move-in costs of both properties.

Rental Property

The monthly rent is $1,350.

The move-in costs are:

First month = $1,350

Last month = $1,350

Security deposit = 55% of one month's rent

[tex]\Rightarrow\frac{55}{100}\times1350=742.5[/tex]

Therefore, the move-in cost is:

[tex]\Rightarrow1350+1350+742.5=3442.5[/tex]

Purchase Property

The purchase price is $195,450.

The move-in costs are:

Down payment of 18% of purchase price:

[tex]\Rightarrow\frac{18}{100}\times195450=35181[/tex]

Closing costs of 2.1% of purchase price:

[tex]\Rightarrow\frac{2.1}{100}\times195450=4104.45[/tex]

Therefore, the move-in cost is:

[tex]\Rightarrow35181+4104.45=39285.45[/tex]

Difference in Total Move-In Cost

This is calculated to be:

[tex]\Rightarrow39285.45-3442.5=35842.95[/tex]

ANSWER

The difference in total move-in cost is $35,842.95

According to Debt.org the average household has $7,281 in credit card debt. Estimate how much interest the average household accumulates over the course of 1 year. We are going to assume the APR is 16.99%.

Answers

In order to estimate the interest the average househould accumulates in 1 year, you use the following formula:

A = Prt

where P is the initial credit card debt ($7,281), r is the interest rate per period (16.99%) and t is the number of time periods. In this case the value of r is given by the APR, then, there is one period of 1 year.

To use the formula it is necessary to express 16.99% as 0.1699. Thus, you have:

I = 7,281 x 0.1699 x 1

I = 1,237.04

Hence, the interest accumulated is of $1,234.04

Owners of a recreation area are filling a small pond with water. They are adding water at a rate of 29 L per minute. There are 400 L in the pond to start. Let W represent the total amount of water in the pond (in liters) and let T represent the total number of minutes that water has been added.Write an equation relating W to T. Then use this equation to find the total amount of water after 13 minutes.Equation : Total amount of water after 13 minutes : liters

Answers

In this problem, we have a linear equation of the form

W=mT+b ----> equation in slope-intercept form

where

m is the unit rate or slope of the linear equation

m=29 L/min ----> given

b is the initial value

b=400 L ----> given

substitute

W=29T+400 -------> equation relating W to T.

For T=13 min

substitute

W=29(13)+400

W=777 L

the total amount of water after 13 minutes is 777 L

Using a graphing calculator to find local extrema of a polynomial function

Answers

The given function is:

[tex]f(x)=3x^4-5x^3-4x^2+5x-2[/tex]

By using a graphing calculator, we found that the local maximum is located at:

x=0.41, then f(0.41)=-0.88

The answer is (0.41, -0.88)

Find the indicated function given f(x)=2x^2+1 and g(x)=3x-5. When typing your answer if you have an exponent then use the carrot key ^ by pressing SHIFT and 6. Type your simplified answers in descending powers of x an do not include any spaces between your characters.f(g(2))=Answerf(g(x))=Answerg(f(x))=Answer (g \circ g)(x) =Answer (f \circ f)(-2) =Answer

Answers

Given the functions

[tex]\begin{gathered} f(x)=2x^2+1 \\ g(x)=3x-5 \end{gathered}[/tex]

1) To find f(g(2))

[tex]\begin{gathered} f(g(x))=2(3x-5)^2+1 \\ f(g(x))=2(9x^2-15x-15x+25)+1=2(9x^2-30x+25)+1 \\ f(g(x))=18x^2-60x+50+1=18x^2-60x+51 \\ f(g(2))=18(2)^2-60(2)+51=18(4)-120+51 \\ f(g(2))=72-120+51=3 \\ f(g(2))=3 \end{gathered}[/tex]

Hence, f(g(2)) = 3

2) To find f(g(x))

[tex]\begin{gathered} f(g(x))=2(3x-5)^2+1 \\ f(g(x))=2(9x^2-15x-15x+25)+1=2(9x^2-30x+25)+1 \\ f(g(x))=18x^2-60x+50+1=18x^2-60x+51 \\ f(g(x))=18x^2-60x+51 \end{gathered}[/tex]

Hence, f(g(x)) = 18x²-60x+51

3) To find g(f(x))

[tex]\begin{gathered} g(f(x))=3(2x^2+1)-5 \\ g(f(x))=6x^2+3-5=6x^2-2 \\ g(f(x))=6x^2-2 \end{gathered}[/tex]

Hence, g(f(x)) = 6x²-2

4) To find (gog)(x)

[tex]\begin{gathered} (g\circ g)(x)=3(3x-5)-5=9x-15-5=9x-20 \\ (g\circ g)(x)=9x-20 \end{gathered}[/tex]

can some one clarify this question, i think ik the answer but i need some elses opinion Find m

Answers

hello

to solve this question, we simply need to add two quadrants that make up mto get m[tex]\begin{gathered} m<\text{WYV}=60^0 \\ m<\text{VYU}=85^0 \end{gathered}[/tex][tex]\begin{gathered} m<\text{WYU}=<\text{WYV}+m<\text{VYU} \\ m<\text{WYU}=60^0+85^0 \\ m<\text{WYU}=145^0 \end{gathered}[/tex]from the calculations above, the value of m

1. Identify the vertex (locator point) of the above parabola2 po(1,2)(3,0)(3,0)(2,1)2. Identify the vertex from the quadratic function y=-5(x-6) 2+82 point

Answers

Answer:

(2,1)

Step-by-step explanation:

The vertex of a parabola is it's highest point(if it is concave down), or it's lowest point, if it's concave up.

In this question:

It's concave down, so the vertex is the highest point.

It happens when x = 2, at which y = 1.

So the vertex is the point (2,1)

An ordinary (Pair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in successionand that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of andereCompute the probability of each of the following svents.Event A: The sum is greater than 7.Event B: The sum is divisible by 3 or 6 (or both).Write your answers as fractions

Answers

[tex]P(A)=\frac{5}{12},P(B)=\frac{1}{3}[/tex]

1) We are going to tackle this question starting with the total outcomes of dice rolled twice in succession.

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

So we can see that there are 36 possibilities.

2) Let's examine the events.

a) P (>7)

Let's bold the combinations of outcomes whose sum is greater than 7

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

So, we can see that there are 15 favorable outcomes.

Now, we can find the Probability of rolling the dice twice and get a sum greater than 7:

[tex]P(A)=\frac{15}{36}=\frac{5}{12}[/tex]

b) Now, for the other event: The sum is divisible by 3 or 6, or both:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Hence, the favorable outcomes are: 12

So now, let's find the probability of getting a sum that way:

[tex]P(B)=\frac{12}{36}=\frac{1}{3}[/tex]

Which angles are adjacent to <2? Select all that apply.

Answers

[tex]\begin{gathered} \text{adjacent angles to <2 are the ones that are on his "sides" thus the adjacent angles to <2 are} \\ <3\text{ and <1} \end{gathered}[/tex]

Jan and her brother mel go to different schools. Jan goes 6 kilometer east from home. Mel goes 8 kilometer north. How many kilometer apart are their schools.

Answers

Jan goes 6 km east from her home, and Mel goes 8 km north from the same home. So we need to find the distance between both schools, and we do such using the Pithagorean theorem, since we are in the presence of a right angle triangle for which we know the two legs, and need to find the measure of the hypotenuse.

I am going to represent the problem with a diagram below, so you see the right angle triangle I am talking about.

So the two legs are represented by the distances each student travels, and the segment in red is the distance between the schools which appears as the HYPOTENUSE of the right angle triangle.

Therefore, we use the Pythagoras theorem for the hypotenuse:

[tex]\begin{gathered} \text{Hypotenuse}=\sqrt[]{leg1^2+\text{leg}2^2} \\ \text{hypotenuse}=\sqrt[]{6^2+8^2} \\ \text{Hypotenuse}=\sqrt[]{36+64} \\ \text{Hypotenuse}=\sqrt[]{100}=10 \end{gathered}[/tex]

Therefore, the distance between the schools is 10 km

3/4 square foot in 1/2 hour what is the unit rate as mixed number

Answers

Answer:

[tex]\text{1 }\frac{1}{2}[/tex]

Explanation:

The unit rate is:

3/4 divided by 1/2

[tex]\begin{gathered} \frac{3}{4}\times\frac{2}{1} \\ \\ =\frac{3}{2} \end{gathered}[/tex]

As a mixed fraction, it is

[tex]\text{1 }\frac{1}{2}[/tex]

Solving triangles using the law of cosines . Find m

Answers

The law of cosines is defined as follows:

[tex]a^2=b^2+c^2-2bc\cos A[/tex]

For the given triangle

a=AC=8

b=AB=14

c=BC=11

∠A=∠B=?

-Replace the lengths of the sides on the expression

[tex]8^2=14^2+11^2-2\cdot14\cdot11\cdot\cos B[/tex]

-Solve the exponents and the multiplication

[tex]\begin{gathered} 64=196+121-308\cos B \\ 64=317-308\cos B \end{gathered}[/tex]

-Pass 317 to the left side of the expression by applying the opposite operation to both sides of it

[tex]\begin{gathered} 64-317=317-317-308\cos B \\ -253=-308\cos B \end{gathered}[/tex]

-Divide both sides by -308

[tex]\begin{gathered} -\frac{253}{-308}=-\frac{308\cos B}{-308} \\ \frac{23}{28}=\cos B \end{gathered}[/tex]

-Apply the inverse cosine to both sides of the expression to determine the measure of ∠B

[tex]\begin{gathered} \cos ^{-1}\frac{23}{28}=\cos ^{-1}(\cos B) \\ 34.77º=B \end{gathered}[/tex]

The measure of ∠B is 34.77º

A 5p coin weighs 4.2g. Approximately, how much will one million pounds worth of 5p pieces
weigh?

Answers

Answer:

It would weight 840,000g

Step-by-step explanation:

1,000,000 ÷ 5

= 200,000

= 200,000 × 4.2

= 840,000

My name is nessalovetrillo i am prepping and studying to test out of my algebra class this is for a study guide Please see attached picture

Answers

Given:

S={(5,6),(-2,-9),(-9,6)}

To find the domain and range:

The domain is,

{-9, -2, 5}

The range is,

{-9, 6}

Calculate the amount of money that was loaned at 4.00% per annum for 2 years if the simple interest charged was $1,240.00.

Answers

Given:-

Simple intrest is $1240. Rate is 4.00%. Time is 2 years.

To find:-

The principal amount.

The formula which relates Simple intrest, Rate, Time and Principal amount is,

[tex]I=prt[/tex]

So from this the formula for p is,

[tex]p=\frac{I}{rt}[/tex]

Subsituting the known values. we get,

[tex]\begin{gathered} p=\frac{I}{rt} \\ p=\frac{1240}{0.04\times2} \\ p=\frac{1240}{0.08} \\ p=\frac{124000}{8} \end{gathered}[/tex]

By simplifying the above equation. we get the value of p,

[tex]\begin{gathered} p=\frac{124000}{8} \\ p=\frac{31000}{2} \\ p=15500 \end{gathered}[/tex]

So the principle amount value is 15500.

Jo-o/checkpoint scatter plotsA2018161412Paw size (centimeters)10642X1b2030405066708090100Height (centimeters)Does this scatter plot show a positive association, a negative association, or no association?positive associationnegative associationno association

Answers

A scatter plot shows the association between two variables.

If the variables tend to increase and decrease together, the association is positive. If one variable tends to increase as the other decreases, the association is negative. If there is no pattern, the association is zero.

From the graph we notice that in this case both variables increcase together, therefore the scatter plot has a positive association.

Tony is a hiring director at a large tech company in Chicago, and he gets hundreds of resumes each week. How long does Tony MOST likely spend looking over each resume?30 seconds50 seconds3 minutes30 minutes

Answers

The time needed to look over the resumes depends on how many papers is the resume

But it is convenient to have a speed looking on each one

so, the answe will be 50 seconds

For an outdoor concert by the city orchestra, concert organizers estimate that 11,000 people will attend if it is not raining. If it is raining, concert organizers estimatethat 7000 people will attend. On the day of the concert, meteorologists predict a 60% chance of rain. Determine the expected number of people who will attend thisconcert

Answers

Step 1

Given;

For an outdoor concert by the city orchestra, concert organizers estimate that 11,000 people will attend if it's not raining.

If it is raining, concert organizers estimate 7000 people will attend.

On the day of the concert, meteorologists predict a 60% chance of rain.

Step 2

Given that the probability of having rain is 60%

[tex]Pr(rain)=\frac{60}{100}=0.6[/tex]

So the probability of not having rain is;

[tex]\begin{gathered} Pr(rain)+Pr(no\text{ rain\rparen=1} \\ Pr(no\text{ rain\rparen=1-Pr\lparen rain\rparen} \\ Pr(no\text{ rain\rparen=1-0.6=0.4} \end{gathered}[/tex]

Step 3

Now, the expected number of people who will attend the concert will be:

=(probability of not having rain x number of expected guests when it does not rain) + (probability of having rain x number of expected guests when rains)

[tex]\begin{gathered} Pr(expected\text{ number of peope\rparen=\lparen0.4}\times11000)+(0.6\times7000) \\ Pr(expected\text{ number of peope\rparen=4400+4200=8600} \end{gathered}[/tex]

Answer; So, the expected number of people who will attend the concert is 8600

P(x) =x and q(x) = x-1Given:minimum x and Maximum x: -9.4 and 9.4minimum y and maximum y: -6.2 and 6.2Using the rational function [y=P(x)/q(x)], draw a graph and answer the following: a) what are the zeroes?b) are there any asymptotes? c) what is the domain and range for this function?d) it it a continuous function?e) are there any values of y= f(x)/g(x) that are undefined? Explain

Answers

we have the following function

[tex]\frac{p(x)}{g(x)}=\frac{x}{x\text{ -1}}[/tex]

where x is between -9.4 and 9.4 and y is between -6.2 and 6.2.

We will first draw the function

from the graph, we can see that the zeroes are all values of x for which the graph crosses the x -axis

In this case, we see that that the only zero is at x=0.

Now, we have that the asymptotes are lines to which the graph of the function get really close to. On one side, we can see that as x goes to infinity or minus infinity, the values of the function get really close to 1. So the graph has a horizontal asymptote at y=1. Also, we can see that as x gets really close to 1, the graph gets really close to the vertical line x=1. So the graph has a vertical asymptote at x=1.

Recall that the domain of a function is the set of values of x for which the function is defined. From our graph, we can see that graph is not defined when x=1. So the domain of the function is the set of real numbers except x=1. Now, recall that the range of the function is the set of y values of the graph. From the picture we can see that the graph has a y coordinate for every value of y except for y=1. So, this means that the range of the function is the set of real numbers except y=1.

From the graph, we can see that we cannot draw the graph having a continous drawing. That is, imagine we take a pencil and start on one point on the graph on the left side. We can draw the whole graph on the left side, but we cannot draw the graph on the right side without lifting the pencil up. As we have to "lift the pencil up" this means that the graph is not continous

Finally note that as we have a vertical asymptote at x=1 and horizontal asymptote at y=1 we have that when y is 1 or x is 1, the function y=f(x)/g(x) is undefined

The table shows the fraction of students from differentgrade levels who are in favor of adding new items tothe lunch menu at their school. Which list shows the grade levels in order from the greatest fraction of students to the least fraction of students ?

Answers

First, write all the fractions using the same denominator. To do so, find the least common multiple of all denominatos. The denominators are:

[tex]50,20,25,75,5[/tex]

The least common multiple of all those numbers is 300.

Use 300 as a common denominator for all fractions to be able to compare their values.

5th grade

[tex]\frac{33}{50}=\frac{33\times6}{50\times6}=\frac{198}{300}[/tex]

6th grade

[tex]\frac{13}{20}=\frac{13\times15}{20\times15}=\frac{195}{300}[/tex]

7th grade

[tex]\frac{18}{25}=\frac{18\times12}{25\times12}=\frac{216}{300}[/tex]

8th grade

[tex]\frac{51}{75}=\frac{51\times4}{75\times4}=\frac{204}{300}[/tex]

9th grade

[tex]\frac{3}{5}=\frac{3\times60}{5\times60}=\frac{180}{300}[/tex]

Now, we can compare the numerators to list the fraction from greatest to lowest:

[tex]\begin{gathered} \frac{216}{300}>\frac{204}{300}>\frac{198}{300}>\frac{195}{300}>\frac{180}{300} \\ \Leftrightarrow\frac{18}{25}>\frac{51}{75}>\frac{33}{50}>\frac{13}{20}>\frac{3}{5} \\ \Leftrightarrow7th\text{ grade}>8th\text{ grade}>5th\text{ grade}>6th\text{ grade}>9th\text{ grade} \end{gathered}[/tex]

Therefore, the list of grade levels in order from the greatest fraction of students to the least fraction of students, is:

7th grade (18/25)

8th grade (51/75)

5th grade (33/50)

6th grade (13/20)

9th grade (3/5)

are figures A and B congruent? explain your reason

Answers

[tex]\begin{gathered} \text{The size of both the figure are not same,} \\ So,\text{ the Figure A and figure B are not cogruent.} \end{gathered}[/tex]

8.1 km to miles and feet

Answers

Given

[tex]8.1\operatorname{km}[/tex]

It should be noted that

[tex]\begin{gathered} 1\operatorname{km}=0.621371miles \\ 1\text{mile}=5280\text{feet} \end{gathered}[/tex][tex]\begin{gathered} \text{convert 8.1km to miles} \\ 1\operatorname{km}=0.621371\text{miles} \\ 8.1\operatorname{km}=8.1\times0.621371 \\ 8.1\operatorname{km}=5.0331051\text{miles} \end{gathered}[/tex][tex]\begin{gathered} 8.1\operatorname{km}=5\text{miles}+0.0331051\text{miles} \\ \text{convert 0.0331051miles to fe}et \\ 1\text{miles}=5280ft \\ 0.0331051\text{miles}=0.0331051\times5280feet \\ 0.0331051\text{miles}=174.79feet \end{gathered}[/tex]

Hence, 8.1km is 5 miles and 174.79 feet

solve the following system of inequalities graphically on the set of axes below?witch of the coordinates points would be in the solution set

Answers

Don’t cheat bro learn ur self ahahaha

what is the image of 2,10 after a dilation by a scale factor of 1/2 centered at the origin

Answers

A dilation is given by:

[tex](x,y)\rightarrow(kx.ky)[/tex]

where k is the scale factor.

In this case we have:

[tex](2,10)\rightarrow(\frac{1}{2}\cdot2,\frac{1}{2}\cdot10)=(1,5)[/tex]

Therefore the image is the point:

[tex](1,5)[/tex]

Please help I need by today only questions 5 and 6 need to show work

Answers

Part a: Pot the points X, Y and Z are obtained on graph.

Part b: Distances; XY = 3, YZ = 5 and XZ = √34 units.

Part c: Measure of angles; X = 59.04 degrees and Z = 30.96 degrees.

What is termed as the Pythagorean theorem?The Pythagorean theorem states that the sum of a squares just on legs of a right triangle equals the square just on hypotenuse.

For the given question, Triangle XYZ with the vertexes are given.

Part a: Pot the points.

X = (6, 6)

Y = (6, 3)

Z = (1, 3)

The points on the graph are plotted.

Part b: Distances;

XY = 6 - 3 = 3 units

YZ = 6 - 1 = 5 units

XZ , use Pythagorean theorem.

XZ² = XY² + YZ²

Put the values.

XZ² = 3² + 5²

XZ² = 9 + 25

XZ² = 34

XZ = √34 units.

Part c: Measure of angles.

In right triangle XYZ

cos X = XY/XZ

cos X = 3/ √34

X = 59.04 degrees.

cos Z = ZY/XZ

cos Z = 5/ √34

Z = 30.96 degrees.

Thus, the value of the triangle are obtained.

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What is the value of x? A pair of intersecting lines is shown. The angle above the point of intersection is labeled left parenthesis 7 x minus 8 right parenthesis degrees. The angle directly opposite below the point of intersection is labeled left parenthesis 6 x plus 11 right parenthesis degrees. (1 point)
A –19
B 125
C 19
D 55

Answers

The value of x in the angles is 19.

How to find angles in intersecting lines?

When lines intersect, angle relationships are formed such as vertically opposite angles, adjacent angles etc.

Therefore, let's find the value of x in the intersecting lines.

Hence,

7x - 8 = 6x + 11 (vertically opposite angles)

Vertically opposite angles are congruent and they share the same vertex point.

Hence,

7x - 8 = 6x + 11

subtract 6x from both sides of the equation

7x - 8 = 6x + 11

7x - 6x - 8 = 6x - 6x + 11

x - 8 = 11

add 8 to both sides of the equation

x - 8 + 8 = 11 + 8

x = 19

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